Skip to main content

bugs - What is the syntax to conditionally suppress plotting in version 9?


Bug introduced in 9.0 and fixed in 9.0.1





In Mathematica 7 it is very easy to conditionally suppress plotting of individual lines using If:


Plot[{
If[x^2 < 2, x^2],
If[Exp[x] > x^2, Exp[x]],
If[False, x] (* check recommended by Rahul *)
},
{x, -2, 2},
PlotStyle -> Thick, Frame -> True]


enter image description here


Or more verbosely using Piecewise and Indeterminate:


Plot[{
Piecewise[{{x^2, x^2 < 2}}, Indeterminate],
Piecewise[{{Exp[x], Exp[x] > x^2}}, Indeterminate],
Piecewise[{{x, False}}, Indeterminate] (* check recommended by Rahul *)
},
{x, -2, 2},
PlotStyle -> Thick, Frame -> True]


enter image description here


It is reported that neither method works in version 9.0.0 (at least on OSX.) Furthermore it is reported that my attempt using ConditionalExpression also fails:


Plot[{
ConditionalExpression[x^2, x^2 < 2],
ConditionalExpression[Exp[x], Exp[x] > x^2]
},
{x, -2, 2},
PlotStyle -> Thick, Frame -> True]

Plotting a zero is reported to "work" but that is hardly a solution:



Plot[{
Piecewise[{{x^2, x^2 < 2}}],
Piecewise[{{Exp[x], Exp[x] > x^2}}]
},
{x, -2, 2},
PlotStyle -> Thick, Frame -> True]

enter image description here


1. Is this indeed a bug in version 9.0.0?


2. Is there a workaround for the affected systems?




Answer



Since this bug seems to be tied directly to the appearance of Indeterminate as the only available function value in the plot range, it could perhaps be considered a work-around to replace Indeterminate by another "quantity" that behaves the same way as Indeterminate but doesn't cause the whole display of all other functions to be suppressed.


I tried the following, and it works on version 9.0.0:


Plot[{Piecewise[{{x^2,x^2<2}},Indeterminate],
Piecewise[{{Exp[x],Exp[x]>x^2}},Indeterminate],
Piecewise[{{x,False}},I] (*modified check recommended by Rahul*)},
{x,-2,2},
PlotStyle->Thick,Frame->True]

Here, I used the imaginary unit I to produce the same effect as Indeterminate, and the remaining plots do still get displayed.



To make this more general, maybe one can use a replacement rule like this:


Plot[Evaluate[{Piecewise[{{x^2, x^2 < 2}}, Indeterminate], 
Piecewise[{{Exp[x], Exp[x] > x^2}}, Indeterminate],
Piecewise[{{x, False}},
Indeterminate] (*modified check recommended by Rahul*)} /.
Indeterminate -> I], {x, -2, 2}, PlotStyle -> Thick, Frame -> True]

Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...